In a triangle \( \mathrm{ABC}, \mathrm{E} \) is the mid-point of median AD. Show that \( \operatorname{ar}(\mathrm{BED})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC}) \).
Given:
In a triangle \( \mathrm{ABC}, \mathrm{E} \) is the mid-point of median AD.
To do:
We have to show that \( \operatorname{ar}(\mathrm{BED})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC}) \).
Solution:
![](/assets/questions/media/153848-64458-1659793499.jpg)
We know that,
A median divides a triangle into two triangles of equal areas.
This implies,
$\operatorname{ar}(\triangle \mathrm{ABD})=\operatorname{ar}(\triangle \mathrm{ADC})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABC})$............(i)
In $\triangle \mathrm{ABD}$
$\mathrm{BE}$ is the median.
This implies,
$\operatorname{ar}(\triangle \mathrm{BED})=\operatorname{ar}(\triangle \mathrm{BAE})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABD})$
$\Rightarrow \operatorname{ar}(\triangle \mathrm{BED})= \frac{1}{2}[\frac{1}{2}ar(\triangle \mathrm{ABC})]$ [From (i)]
$\operatorname{ar}(\triangle \mathrm{BED})=\frac{1}{4} \operatorname{ar}(\triangle \mathrm{ABC})$
Hence proved.
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